Help!! it says: Determine if f(x)=4x^(2/6) is even, odd, or neither?

Answer 1

x^(2/6) is not necessarily the same as x^(1/3)

x^(2/6) is defined as the 6th root of x^2.
If we are only interested in real values of f(x), and if we define the 6th root as +ve only, then the function is even.

If we define the 6th root any other way, f(x) is multivalued. If the 6th root is defined as +ve and -ve, then the graph is actually both even and odd.

Answer 2

See answer to previous problem about 4x^(2/3)

Answer 3

Is this f(x) = 4x^(1/3)?

A function is even iff f(-x) = f(x), and odd iff f(-x) = -f(x).
For your function, f(-x) = 4(-x)^(1/3) = 4*((-1)^(1/3))*x^(1/3) =
4*(-1)*x^(1/3) = -4x^(1/3) = -f(x). Therefore, this function is odd.

Answer 4

even